Example Problem with Complete Solution

8D-2 : Entropy Generation and Lost Work for a Compressor with Heat Losses 7 pts
The outer surface of a air compressor is at an average temperature of 485 K and the surroundings are at 300 K. The operating parameters for the compressor are given in the figure below.
a.) Calculate the internal, external and total entropy generation rates for the compressor in kW/K.
b.) Calculate the internal, external and total lost work for the compressor in kW.
c.) Show that the total lost work determined in part (b) is the difference between the power requirement of a completely reversible compressor that accomplishes the same change in state of the air stream and the actual work for the 
real compressor.
Read : Equations for evaluating total and external entropy generation can be obtained from Lesson 8D.  You could just take the difference between these to determine the internal entropy generation.  But it is useful to understand how the position of the system boundary can be manipulated to directly yield an equation for the internal entropy generation.
The keys are to use the 1st Law to determine Q and to use the 2nd Gibbs Equation and the Ideal Gas Entropy Function to evaluate DS.
Once the entropy generation rates are known, it is easy to determine the lost work rates.
In order to verify the definition of lost work, we must determine the amount of work a reversible compressor would require to accomplish the same compression process.  Qrev must be different from Qact. The key to this part of the problem is that total entropy generation rate for the reversible compressor must be zero.
Given: P1 96 kPa T1 310 K
P2 1200 kPa T2 640 K
m 1.30 kg/s Tsurf 485 K
Ws,act -555 kW Tsurr 300 K
Find: a.) Sgen,int ??? kW/K b.) Wlost,int ??? kW
Sgen,ext ??? kW/K Wlost,ext ??? kW
Sgen,tot ??? kW/K Wlost,tot ??? kW
c.) Show that :
Eqn 1
Assumptions: 1 - The surface temperature of the compressor remains constant. Therefore, it can be treated as a thermal reservoir.
2 - The compressor operates at steady-state with negligible changes in kinetic and potential energies.
3 - Air behaves as an ideal gas.
Diagram: The diagram in the problem statement is adequate.
Equations / Data / Solve:
Part a.) We can use the following equations to evaluate the internal and total entropy generation rates.
Eqn 2
If we place the system boundary far from the surface of the compressor, then ALL of the irreversibilities are inside the system because the temperature at the system boundary is the same as the temperature of the surroundings. So, heat exchange between this big system and the surroundings is reversible.
This explains why using Tsurr in Eqn 2 tells us that we are computing the TOTAL entropy generation rate.  Because with this big system, there are no irreversibilities OUTSIDE the system boundary.  Notice that heat transfer from this system does occur, but it occurs with zero temperature driving force, so it is reversible !
Eqn 3
Using our usual system boundary, right at the surface of the compressor, heat exchange with the surroundings is irreversible. So, by using Eqn 3, with Tsurf instead of Tsurr, we have excluded the external irreversibility due to heat transfer through a finite temperature difference.  So, this entropy generation equation gives us the INTERNAL entropy generation only !
The external entropy generation rate is equal to the rate at which the entropy of the universe (compressor and surroundings) changes due only to the heat transfer from the compressor to the surroundings.
Eqn 4
The minus sign appears in Eqn 4 because the sign of Q from the perspective of the system is negative, but from the perspective of the surroundings, Q > 0 because heat is entering the surroundings !
Eqn 4 can be rearranged to give us :
Eqn 5
Let's begin by determing DS for the working fluid using the 2nd Gibbs Equation and the Ideal Gas Property Table for air.
Eqn 6
So1 0.038914 kJ/kg-K So2 0.78826 kJ/kg-K
R 8.314 J/mole-K
MW 28.97 g/mole DS 0.02450 kJ/kg-K
Next, we need to evaluate Q, the rate of heat loss from the compressor. We can do this using the 1st Law for open systems operating at steady-state with negligible changes in kinetic and potential energies.
Eqn 7
We can solve Eqn 7 for Q :
Eqn 8
Now, we can use the Ideal Gas Property Table for air to evaluate H1 and H2.
Ho1 97.396 kJ/kg Ho2 439.98 kJ/kg
Plug values back into Eqn 7 to evaluate Q : Q -109.64 kW
Now, we are able to plug values back into Eqns 2, 3 & 5 to evaluate the entropy generation rates.
Sgen,int 0.25791 kW/K
Sgen,ext 0.13941 kW/K Sgen,tot 0.39731 kW/K
Double check your calculations using :
Eqn 9
Part b.) Once we have completed part (a), part (b) is a straightforward application of the relationship between lost work and entropy generation.
Eqn 10
Eqn 10 applies for internal, external and total entropy generation and lost work.
Wlost,int 77.37 kW
Wlost,ext 41.82 kW Wlost,tot 119.19 kW
Part c.) Here we must verify that our answer from part (b) is consistent with the defintion of lost work.
Eqn 11
In order to do this, we must evaluate WS,rev.
The key to doing this is to understand that the reversible process still operates between state 1 and state 2.
The values of Q and WS are different from those for the real compressor.
The fact that ties this part of the problem together and allows us to determine Q and WS for the reversible compressor is that :
Eqn 12
We can use this fact and solve Eqn 2 for Qrev :
Eqn 13
Qrev 9.55 kW
Next, we can apply the 1st Law to the reversible compressor to evaluate WS,rev.
We can solve Eqn 7 for WS,rev.  The result is :
Eqn 14
Plugging values into Eqn 14 yields : WS,rev -435.81 kW
Finally, put values into the right-hand side of Eqn 11 :
119.19 kW
This matches our result for lost work from part (b).  So, we have confirmed the relationship between reversible work, actual work and lost work.
Verify: None of the assumptions made in this problem solution can be verified.
Answers : a.) Sgen,int 0.258 kW/K b.) Wlost,int 77.4 kW
Sgen,ext 0.139 kW/K Wlost,ext 41.8 kW
Sgen,tot 0.397 kW/K Wlost,tot 119.2 kW
c.) WS,rev - WS,act = 119.2 kW